let x = .9999..., then 10x=9.999... 10x-x=9x, 9x=9, x= l, therefore 1=.9999...?

Question

comments? This was from a math class, it seems to prove that 1 = .99999...

Answer 1

No need to comment on your proof. It's excellent.

Now to comment on the others:
• Eulercrosser is as bright as ever.
• epg3 is an idiot. 0.9999... (repeating forever) is equal to one, not less than one as the first line in your so-called proof claims.
• asaaiki is an idiot. How can you arbitrarily use significant figures for a number with infinitely many of them?
• snowy is an idiot. There's absolutely nothing wrong in saying 10x - x = 9x, or that 9.9999... (repeating forever) - 0.9999... (repreating forever) = 9. The "forever" of nine digits following the decimal point all subtract to a difference of 0's forever.
• Dr. Rob explains it well.
• gnusselt is an idiot. Since 0.9999... (repeating forever) = 1, so 8.9999... (repeating forever) = 8 + 1, which does indeed equal 9.
• hi_patia is right on the money (and is also swiping her name from one of my favorite teachers in history).
• prune scores 100% for his proof.
• Mathematician, nicely done as always.
• Dr. blahb31's final paragraph sums it up nicely for the idiots.

Answer 2

This is flawed. Sorry to contradict the so-confident answer calling everyone else idiots.

0.9999..., being inifinitely close to 1, is still not equal to 1. How you can justify them being the same number is beyond me.

Your proof starts with x = .9999. Then later says that 9x = 9, when it should be 8.9999...

Why? If 9x = 9, then x = 1. Not 0.9999...

That's where the trick is. epg3 was right.

By the way, computers/calculators will often return a number with seemingly recurring digits at the end. In that case it's generally safe to assume the rounded figure is correct, since calculators (and most computer calculation programs) do NOT worry about recurring numbers. They just work with a number of digits (16 usually) and return the result accordingly.

You, yourself, know that 1 is not equal to 0.9999...

That's why you're here. So listen to your own logic, and you'll see who is right.

Answer 3

This is true. From the study of infinite sequences and series, you will find that the geometric series a + ar^2 + ar^3 + ar^4..... =
a/(1-r) if -1 (9/10) + (9/10)(1/10)^2 + (9/10)(1/10)^3 + (9/10)(1/10)^4..... In this case a=9/10 and r = 1/10 which is is between -1 and 1.
Therefore, 0.9999 = (9/10)/(1-1/10) = (9/10)/(9/10) = 1.

Answer 4

You lose a significant digit this way
9.9990-0.9999 = 8.9991, and then it all adds up, but the point here is that you only have 9.999, and that that a trailing zero is a significant digit.

I think this example is used to show the importance of significant digits

correction - this is true only as long as there is a finite number of digits. If it's infinite, I guess 0.999... really equals 1, somehow. Amazing.

Answer 5

That is one of several ways to reason that 0.9999... = 1. Here is another.

0.999.... = 0.9 + 0.09 + 0.009 + ....
= 9*(0.1 + 0.01 + 0.001 + ...)
= 9*(0.1^1 + 0.1^2 + 0.1^3 + ...)
= 9*{0.1/(1-0.1)} (Since |0.1| < 1)
= 9*(1/9)
= 1

or

1/3 = 0.3333....
2/3 = 1/3+1/3 = 0.33333.... + 0.333333.... = 0.666666......
3/3 = 1/3+1/3+1/3 + 0.3333.... + 0.33333..... + 0.333333... = 0.99999......
so 1 = 0.999999....

or

There is no number in between 1 and 0.9999.... Since two different numbers always have a number in between them, they must be the same number.

Those of you above who say otherwise do not understand the point that the 9's repeat FOREVER! Any rational number that does not have a repeating decimal has a representation that does have a repeating decimal (ie 1.25 = 1.249999....).

edit: Some of the people who answered disagree. All I have to say is this: Look at the people who answered, along with their profiles. Mathematician, Eulercrosser, Louise, myself (hopefully not too presumptive of me to include myself)... we all have strong backgrounds in mathematics on Yahoo! Answers and in real life. We are all saying the same thing. 0.999...=1. Period.

Answer 6

Actually this is a common trick with infinite numbers.
0.9999... = 1 - 0.000000000...001
The last value tends toward zero otherwise you have a finite number of 9. So 0.9999... = 1 - ~0 and yes it is equivalent to 1.

More fun:
1/3 = 0.33333333333...
3 * 1/3 = 0.99999999999 = 1

Answer 7

let x = .9999..., then 10x=9.999... 10x-x=9x, 9x=9, x= l, therefore 1=.9999...?

U r wrong

When x is in decimal, 10x-x=9x euation is not valid ... cannot hold

u have to keep continue

10x=9.999
10x-x=9.999-x

Answer 8

It does prove that 1=.99999......., because its true.

Think about this:
1/9=.111111......
2/9=.222222.......
3/9=.333333.....
.
.
.
8/9=.88888.....
9/9=.99999.....

9/9=1=.9999.....

The secret is all in those little dots that show the sequence goes on for ever.

Answer 9

x < 1, 10x < 10, 9x < 9, 1 != 0.9999....

Answer 10

forgot a 9 in the subtraction .999 - .9999